racket head size

I think there were discussions on this in the past - but I can't find a definitive answer.

When racket specs are published, and they say 'racket head: 98 sq' - how exact is that number? I'm asking since I was doing some measurements on my own (I know, I have too much time) and the area I was getting was consistently bigger than what the advertised number was.

Like advertised 95 was more of 99 sq, 98 was around 105 and so on. (tried with few head rackets).

I have only one Head racquet (Head 660 Lady) which is nominally 660 square centimeters (102 sq. inches).

According to my measures (inside the hoop), the head area is 661 square centimeters, which is practically as advertised (I calculated the area of "ideal" ellipse, while the actual shape is slightly like the teardrop, so that may cause slight variations).

For Wilson BLX Pro Open I calculated 636/98, slightly less than advertised 645/100.

Take a piece of paper larger than the head of the racquet. Measure it and determine its area in cm (length x width). Weigh it on a small scale, in grams for ease of calculation but you can use the imperial system if you are a nostalgic. Divide weight by surface. Save the result. Draw the internal circumference of the head of the racquet. Cut out the oval and weigh it. Divide the result by the number saved previously. The result is the area in square centimeters (or whatever units you used). Simple enough.

Take a piece of paper larger than the head of the racquet. Measure it and determine its area in cm (length x width). Weigh it on a small scale, in grams for ease of calculation but you can use the imperial system if you are a nostalgic. Divide weight by surface. Save the result. Draw the internal circumference of the head of the racquet. Cut out the oval and weigh it. Divide the result by the number saved previously. The result is the area in square centimeters (or whatever units you used). Simple enough.

Most modern rackets (not the tear-shaped ones of the past) have heads which are approximately ellipses. Measure the "length" a and "width" b in inches of the head at the maximum span points. The area is (0.785*a*b) square inches.

its very difficult to calculate a racket head area on your own since its not a perfect ellipse or circle so you might be getting wrong numbers because you are not using the correct calculations.

Click to expand...

yah, but why do you think I was assuming that a racket head is an ellipse or a circle?
What I did was to take a picture of a racket next to a shape (like a rectangle or a circle) with a known area in sq inches. Imported that into any decent image editor. In the editor I find out how many pixels the racket head area takes. I also find out how many pixels the known-size area takes. Once I have that it is a simple equation to find out the area of the racket head.
Works pretty well. Just need to be careful to take proper pictures and do some image editing magic to make sure that in the editor the pixels are counted properly.
To be even more sure I usually take pictures of the racket next to multiple shapes with known area. That way I can re-verify the measurements and pixels-to-area equations. Since I know the areas of at least two shapes I can verify if the pixel ratio for those two shapes makes sense (as it needs to be the same ratio as the ratio of their areas).

For example, I measured and compared my 4D 200 Tours and PSGTs, both unstrung. The hieght and width of the frames is the same. The PSGTs simoly have a lttle more space in the bottom "corners" and play like a more forgiving version of the 95" 200s.